as you already know, to get the inverse of any expression, we start off by doing a quick switcharoo on the variables, and then solve for "y".
[tex]\bf \stackrel{f(x)}{y}=0.5(3)^x\qquad \stackrel{inverse}{\boxed{x}=0.5(3)^{\boxed{y}}}
\\\\\\
log(x)=log[0.5(3)^y]\implies log(x)=log(0.5)+log[(3)^y]
\\\\\\
log(x)-log(0.5)=log[(3)^y]\implies log(x)-log(0.5)=y\cdot log(3)
\\\\\\
\cfrac{log(x)-log(0.5)}{log(3)}=\stackrel{f^{-1}(x)}{y}\\\\
-------------------------------\\\\
f^{-1}(7)=\cfrac{log(7)-log(0.5)}{log(3)}\implies f^{-1}(7)\approx 2.402173502732879697[/tex]
